The angle theta between vectors vec{a} = hat{ | vedclass

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(B) The angle $	heta$ between two vectors $\vec{a}$ and $\vec{b}$ is given by the formula $\cos 	heta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec

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$\pi - \cos^{-1}\left(\frac{1}{3}\right)$

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(B) The angle $	heta$ between two vectors $\vec{a}$ and $\vec{b}$ is given by the formula $\cos 	heta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}$.Given $\vec{a} = \hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$.The dot product is $\vec{a} \cdot \vec{b} = (1)(1) + (1)(-1) + (-1)(1) = 1 - 1 - 1 = -1$.The magnitudes are $|\vec{a}| = \sqrt{1^2 + 1^2 + (-1)^2} = \sqrt{3}$ and $|\vec{b}| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3}$.Thus,$\cos 	heta = \frac{-1}{\sqrt{3} \cdot \sqrt{3}} = -\frac{1}{3}$.Therefore,$	heta = \cos^{-1}\left(-\frac{1}{3}ight)$.Using the property $\cos^{-1}(-x) = \pi - \cos^{-1}(x)$,we get $	heta = \pi - \cos^{-1}\left(\frac{1}{3}ight)$.Hence,the correct option is $B$.

The angle $\theta$ between vectors $\vec{a} = \hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$ is . . . . . . .

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